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#129884 10/10/05 07:55 PM
Joined: Sep 2003
Posts: 114
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I mostly deal with residential 120V/240V systems. What I'm wondering is the specifics of how electricity flows. I have read some basic texts that try to explain it, and perhaps they do, but I'm still a bit confused. Let me explain how I think it works and any of you can feel free to add to it or correct me.

A pot transformer steps down 6kVA or what ever to 240V with a secondary side tap in the middle to achieve 120V from either leg.
Question: Since the tap is to earth from the transformer, why isn't there a short or something at the tap? Every drawing I've seen shows the secondary winding being connected to in the center. It seems this would still carry a voltage and would short when it is connected to earth.

For 220V dryers, ovens, etc. It is my understanding that the electrons in each leg move back and forth at whatever the amps pull. So there is no direction for the current going from negative to positive because there is no positive as in DC. Does the current get "consumed" or is there a loss due to heat or whatever that is measurable.

For 120V applications, it seems that the current only moves in one direction; since the neutral doesn't have the same potiential as a negative "hot" to carry current because it is tied back to earth at the panel. It seems that the electricity in this example continually moves toward the neutral from the hot, and the voltage is only gained from the upper half of the sine wave created by the AC generator. As opposed to the top and bottom halves of the sine wave creating 240V between them. So, the only current that the neutral carries is the unbalanced load.

Speaking of unbalanced loads, in the case of 120V applications, where is the unbalanced load comming from? Also, why does it vary in amperage?

Any comments that will help clear these issues up would be appreaciated. I'm looking forward to learning as much as I can.

Thanks,
Byron

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Joined: Sep 2005
Posts: 202
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"Question: Since the tap is to earth from the transformer, why isn't there a short or something at the tap?"

Current flow occurs when there is a difference in potential. As long as there is only one grounded point, there is no short. It is actually only a reference point. A transformer could be connected with any one of the three connection points to ground without it being a short AS LONG AS THERE IS ONLY ONE POINT GROUNDED. Of course, the load (house) would have to be wired accordingly.

"So there is no direction for the current going from negative to positive because there is no positive as in DC"

Actually there is. When you refer to AC, you are referring to a field that reverses polarity. There is a negative and positive relationship: it just reverses at 60 times a second. Current is not consumed...current is the flow of electrons. Think of the X1 and X3 terminals of a transformer as reversing polarity at 60 cycles. X1 is postitive in regard to X3, and then X3 is positive in regard to X1. The current goes back and forth at each change. But there is a voltage drop across the load as this happens.

Think of a water driven paddle wheel. The water itself is the current (amps) and the gravitational drop or pressure across the paddles is the voltage. The pressure of the water across the paddles does the work. The water itself is not consumed.

"For 120V applications, it seems that the current only moves in one direction; since the neutral doesn't have the same potiential as a negative "hot" to carry current because it is tied back to earth at the panel. It seems that the electricity in this example continually moves toward the neutral from the hot, and the voltage is only gained from the upper half of the sine wave created by the AC generator. As opposed to the top and bottom halves of the sine wave creating 240V between them. So, the only current that the neutral carries is the unbalanced load"

I really don't understand exactly what you are saying here, other than the last sentence is correct.

"Speaking of unbalanced loads, in the case of 120V applications, where is the unbalanced load comming from? Also, why does it vary in amperage?"

Think again of the transformer with its three connections (X1, X2 or ground, and X3). Let's freeze time at one point in the cycle. If X3 is positive in regard to X1, then X2 is also positive in regard to X1, but negative in regard to X3. It's at a "mid-potential" between the other two.

So in a load connected from X3 to X2, the curent flows from X3 to X2. In a load connected from X2 to X1, (at this same instant in time) the current flows from X2 to X1. If a load is connected across each, the current flows from X3 to X1 and only the imbalance goes inot X2 (as you noted).

By Ohms law, the current through a load is proportional to its voltage and impedance. Lets say X3 to X2 is 120 volts across a 12 ohm resistance.

120 volts/12 ohms = 10 amps

X2 to X1 is 120 volts across 24 ohms.

120 volts / 24 ohms = 5 amps.

In this case 10 amps is leaving X3, with 5 amps returning through X2 and 5 amps returning through X1. So far so good.

But inside the transformer, the winding from X2 to X3 is carrying twice the current that the X1 to X2 half is.

Let's exaggerate a little. If the load between X3 and X1 was 1 ohm, there would be 120 amps flowing out of X3 and 115 going in X2. So now half the transformer has 115 amps on it, and the other half only has 5. Hold this thought for a minute.

A 10 KVA transformer at 240 volts is rated for about 41 amps. Since the internal windings are in series, each can carry 41 amps. Remember that the copper in this winding can only carry 41 amps without exceeding its current rating. If you put all 41 amps onto one leg, that winding is at capacity. But since it is at 120 volts (120 v X 41 amps = 5 Kva), you are only utilizing one half of the total capacity of the transformer. By splitting the load between the two windings, you can utilize the entire transformer.

Probably went off on a tangent or two here, but I hope it helps.

Joined: Sep 2005
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I hate having to correct my own post.

"Let's exaggerate a little. If the load between X3 and X1 was 1 ohm,"

That should have read X3 to X2.

Sorry.....I'm not used to typing that much.

Joined: Feb 2005
Posts: 693
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Byron, how about a simpler explanation first, about a tapped power supply? Forget about AC for now, and concentrate on DC.

Picture the two batteries inside a typical flashlight: One on top of the other, - to +, which places them in series. There is 3 volts between the two remaining terminals.

If you insert a wire between the batteries, you still have the 3-volts end-to-end, but you can also receive 1.5 volts between the center and either end; this is the "neutral".

These two 1.5-v sources can be used as independentl suorces, and Ohm's Law calculations can be applied independently to each one; this is how we profecssionals do it.

You can run a 3-v bulb across the two batteries, or a 1.5-volt bulb between the center tap and either end; you can even run two 1.5-v bulbs and a 3-volt one simultaneously.

If one 1.5-v bulb used 1 amp, and the other used 0.75 amps, one end wire would carry 1 amp, the other would carry 0.75 amps, and the "neutral" wire would carry the difference, 0.25 amps.

If the 3-v bulb used 2 amps, then one end wire would carry 3 amps, the other would carry 2.75 amps, and the "neutral" would still carry the difference, or the same 0.25 amps.

You should be able to translate this center-tapped idea to the 240/120-volt systems you're used to working with. That the system is AC instead of DC is of no relevance here.

The main advantage of AC over DC is the ability to use transformers. Now, as you surmised, a residential-serving utility transformer has a grounded-center-tapped secondary winding.

There is 240 volts between the two end terminals, and 120 between the neutral and either end terminal. Grounding the neutral minimizes the voltage-to-earth potential for safety.

Here's where you ran aground: for line-to-neutral (120-v) loads, each 120-v "circuit" is comprised of a complete sine wave, both the positive and negative halves of the wave.

If you have a 10-amp load on (let's call it) phase A and a 15-amp load on phase B, the 5-amp difference current on the neutral is the result of the 10- and 15-amp currents "clashing".

In other words, anywhere in an AC circuit, the entire full wave flows through all current-carrying conductors; it's the current magnitutes that vary, not the waveform shapes.

If you had two oscilloscopes, one across phase A, and the other across phase B, you'd get images of opposing, apparently-conflicting, waveforms on the neutral at the same time.

The neutral's current would also be a complete sine wave, but of a current level equal to only the difference between the two hots. In theory, the neutral-to-earth voltage is always zero.

Now, in basic-electricity discussions, we ignore conductor (DC) resistances, voltage drops, and capacitive and inductive (AC) reactances. These are taught in next year's class.


Larry Fine
Fine Electric Co.
fineelectricco.com
Joined: Aug 2001
Posts: 7,520
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And remember that the 3-wire DC system Larry is describing with batteries is just a smaller scale version of the way Edison distributed DC power around a neighborhood in the early days of electrical power. (We had such 3-wire DC systems in England until well into the 1950s.)

Joined: Feb 2005
Posts: 693
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Correctamundo, Paul. We have Nikola Tesla and George Westinghnouse to thank for the use of AC today.


Larry Fine
Fine Electric Co.
fineelectricco.com

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